Question

A heat engine uses a diatomic gas in a Brayton cycle. What is the engine's thermal efficiency if the gas volume is halved during the adiabatic compression?

Short answer

The engine's thermal efficiency during adiabatic compression,$\eta =24\%.$

Step-by-step solution

Step: 1 Equating Adiabatic process equation:

Any thermal cycle's efficiency is expressed as

$\eta =1-\frac{T_c}{T_h}$

We know that for an adiabatic process

$p{V}^{\gamma }=\text{const}$

From the first law, we may express pressure in terms of temperature.

$$\begin{gathered} pV=nRT \\ \Rightarrow p=\frac{nRT}{V}\propto T{V}^{-1} \end{gathered}$$

As a result, we have the following for an adiabatic process:

$T{V}^{\gamma -1}=\text{const.}$

Step: 2 Equating:

Let us now calculate a temperature ratio based on the volume ratio:

$T_c{V}_c^{\gamma -1}=T_h{V}_h^{\gamma -1}$

We get that

$\frac{T_c}{T_h}=\frac{V_h^{\gamma -1}}{V_c^{\gamma -1}}={\left(\frac{V_h}{V_c}\right)}^{\gamma -1}.$

Step: 3 Finding the value of Eficiency:

The Efficiency will be,

$\begin{array}{r}\eta =1-{\left(\frac{V_h}{V_c}\right)}^{\gamma -1}\\ \eta =1-{\left(\frac{1}{2}\right)}^{1.40-1}=24\mathrm{\%}.\end{array}$